Question

The value of $g$ at a particular point outside the earth is $10\,m/{s^2}$ . Suppose the earth suddenly shrinks to half its present size. The value of $g$ at that particular point shall now be:
(A) $5\,m/{s^2}$
(B) $15\,m/{s^2}$
(C) $10\,m/{s^2}$
(D) $17\sqrt 2 \,m/{s^2}$

Answer 1

Hint:The gravitational force is the weak force of the attraction offered by the earth. When the object falls due to gravity, it has zero velocity but the movement of the body is due to the acceleration due gravity. If the object is made to move upward, then the external force has to be given against the force of gravity.

Formula used:
The formula of the acceleration due to gravity is given as

$a = \dfrac{{GM}}{{{R^2}}}$

Where $a$ is the acceleration due to gravity, $G$ is the universal gravitational constant, $M$ is the mass of the earth and $R$ is the distance between masses on the surface of the earth.

Complete step by step solution:
It is given that the

Value of the acceleration due to gravity, $a = 10\,m{s^{ - 2}}$

The gravity is the force of attraction that happens between the surface of the earth and the objects on the earth. But this force of attraction is limited only up to a certain height, after that there will be no gravitational force of attraction. Using the formula of the acceleration due to gravity,

$a = \dfrac{{GM}}{{{R^2}}}$

The acceleration due to gravity solely depends on the mass of the earth and the distance between the masses on the surface of the earth. By reducing the size to half, the mass of the earth and the distance between the masses remains the same. Hence the value of the acceleration due to gravity is $10\,m{s^{ - 2}}$ .

Thus the option (C) is correct.

Note:The value of the gravitational constant is the same in all the areas of the universe. The earth’s gravity originates from all of its mass and this causes the pull force all over the body. This gravity is very strong over its surface and very less at its inner core.